Say we want to split our baccarat betting life into endless four-wager spots, anytime registering our W/L ratio by a simple flat betting strategy.
It doesn't matter whether we're betting those four spots consecutively or diluted at various degrees. Let alone which bet selection we would like to use.
Forgetting for now the game asymmetricity, the probability to win or lose all those spots is 1/16 (6.25%), the probability to win at least one wager over four attempts is 15/16 (93.75%).
Easy.
Now say we want to register what happens (by a mere FB placement) after a given not-bet outcome (W or L) had appeared.
The possible results are:
WWWW: +3
WWWL: +1
WWLW: +1
WWLL: -1
WLWW: +1
WLWL: -1
WLLW: -1
WLLL: -3
LLLL: -3
LLLW: -1
LLWL: -1
LLWW: +1
LWLL: -1
LWLW: +1
LWWL: +1
LWWW: +3
Of course the total sum is zero, anyway the symmetrical W/L situations among the 16 possible outcomes are just six (WWLL, WLWL, WLLW, LLWW, LWLW and LWWL).
Math teachs us that no matter which spot we'll decide to bet, any W/L pattern will get the same probability to appear. More specifically that at baccarat every spot wagered will get, itlr, a 50.68/49.32 probability to happen.
In reality the actual card distribution could endorse or not the probability to get, per every four-spots wagered, a symmetrical or asymmetrical situation.
Actually the above considerations reflect a perfect symmetrical 50/50 production, but baccarat is a slight asymmetrical game as itlr B>P.
It may happen that along the shoe we're playing at the slight asymmetricity will endorse a "fictional" simmetricity or, on the other end, increasing a natural asymmetricity.
How can we do to "solve" the problem?
as.
It doesn't matter whether we're betting those four spots consecutively or diluted at various degrees. Let alone which bet selection we would like to use.
Forgetting for now the game asymmetricity, the probability to win or lose all those spots is 1/16 (6.25%), the probability to win at least one wager over four attempts is 15/16 (93.75%).
Easy.
Now say we want to register what happens (by a mere FB placement) after a given not-bet outcome (W or L) had appeared.
The possible results are:
WWWW: +3
WWWL: +1
WWLW: +1
WWLL: -1
WLWW: +1
WLWL: -1
WLLW: -1
WLLL: -3
LLLL: -3
LLLW: -1
LLWL: -1
LLWW: +1
LWLL: -1
LWLW: +1
LWWL: +1
LWWW: +3
Of course the total sum is zero, anyway the symmetrical W/L situations among the 16 possible outcomes are just six (WWLL, WLWL, WLLW, LLWW, LWLW and LWWL).
Math teachs us that no matter which spot we'll decide to bet, any W/L pattern will get the same probability to appear. More specifically that at baccarat every spot wagered will get, itlr, a 50.68/49.32 probability to happen.
In reality the actual card distribution could endorse or not the probability to get, per every four-spots wagered, a symmetrical or asymmetrical situation.
Actually the above considerations reflect a perfect symmetrical 50/50 production, but baccarat is a slight asymmetrical game as itlr B>P.
It may happen that along the shoe we're playing at the slight asymmetricity will endorse a "fictional" simmetricity or, on the other end, increasing a natural asymmetricity.
How can we do to "solve" the problem?
as.